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Homework Help: Thermo question - final mass/entropy generation

  1. Mar 28, 2007 #1
    1. The problem statement, all variables and given/known data
    An insulated and initially empty 0.7m^3 tank is connected by a valve to a line flowing air at 50degreesC and 200kPa. The valve is opened until the tank is filled, and then the valve is closed. The insulation is then removed from the tank, and a 20degreeC heat sink cools the tank until the air inside reaches a temperature of 30degreesC. 1) What is the final mass of air in the tank? 2) How much entropy is generated during the cooling of the tank?

    2. Relevant equations

    3. The attempt at a solution
    1) I tried figuring out the final temperature, and since V and R are constant, I got P2=P1*T2/T1=200kPa*303K/323K=187.62kPa. I then used this P to get m=PV/RT. So m=(187.62*0.7)/(0.287*323)=1.417kg.. Is this right?
    2) Using the first equation, I get 1.004*ln(303/323)-0.287*ln(187.62/200)=-0.0458.. But this is wrong right since it shouldn't be negative?
    Then I try putting it into the second equation, 1.417*-0.0458 - Integral(dQ/303)=S-gen.. but I don't know how to do the Integral and dQ part... any help? THANKS!!
  2. jcsd
  3. Mar 29, 2007 #2
    well since no one else is helping, I'll take a stab, but in the buyer beware category, i know very little about thermo.

    I believe its negative because it is cooler, and altho under less pressure, the temperature difference is dominant. It may be that the question is asking about what entropy must have been generated in the surroundings to accomplish this which would then be >=0.0458 depending on efficiency. integral dQ here becomes deltaQ i believe.
  4. Mar 29, 2007 #3


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    In 1), your equation for m contains a final state pressure and an initial state temperature. You can compute the mass in either the initial state or final state - but you can't mix and match numbers between the two. For 2), why not just do it in terms of the heat capacity at constant volume? Why bring pressure in at all? Oh, it would also be nice if you could be more complete about including units on quantities.
  5. Mar 29, 2007 #4
    Dick and Denver! Thanks so much for responding! I really appreciate it...

    For 1).. I accidentally put in T1! Oops!
    P2 = P1*T2/T1 = 200kPa*303K/323K = 187.62kPa
    So it should be (187.62kPa*0.7m^3)/(0.287kJ/kg-K*303K) = 1.51kg.

    Do you mean the equation s2-s1=C*ln(T2/T1)? I have that down as entropy change for a liquid or solid.
    For an ideal gas (air), I have
    s2-s1 =
    Cp (specific heat) * ln(T2/T1) - R * ln(P2/P1)
    Cv (specific heat) * ln(T2/T1) + R * ln(v2/v1)

    So I used the first one since I have P. Is my logic completely off? :\
  6. Mar 29, 2007 #5


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    Well, either your logic is off, or it's been too long since I've done thermodynamics. But look at your second form. v2=v1. ln(1)=0.
  7. Mar 29, 2007 #6
    OMG Dick! Congrats on 1000 posts! :) :yuck: :!!) :surprised Haha..

    Anyways.. you're right.. I was using the constant pressure equation (boy, do I feel stupid.. although I already do after putting the wrong T in the first post).. ok.. I totally understand now.. I think.. here's my work -

    After I get m = 1.51 kg from part 1 -

    I get s2-s1 = Cv*ln(T2/T1) + R*ln(v2/v1) = 0.717kj/kg-K * ln(303K/323K) + R*0
    = -0.04583 kJ/kg-K

    I also need to get 1Q2. I use 1Q2 = 1W2 + m*(u2-u1) = 0 + 1.51kg*(216.52-230.887) = -21.694 kJ. I got the u2-u1 from interpolating using a chart of ideal-gas properties of air.

    Then I plug them into the equation with s-gen. So
    s-gen = m(s2-s1) - 1Q2/T-source = 1.51kg * -0.04583 kJ/kg-K - (-21.694 kJ)/(20+273.15) = 0.00484kJ/K!!!!

    OMG, I think I did it correctly!!
  8. Mar 29, 2007 #7


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    Good job! Then that was a worthwhile 1000th post. Thanks!
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